3.240 \(\int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx\)

Optimal. Leaf size=170 \[ -\frac{x \text{PolyLog}\left (3,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{i \text{PolyLog}\left (4,-(1-i d) e^{2 i a+2 i b x}\right )}{8 b^3}+\frac{i x^2 \text{PolyLog}\left (2,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)+\frac{1}{12} i b x^4 \]

[Out]

(I/12)*b*x^4 + (x^3*ArcCoth[1 - I*d + d*Tan[a + b*x]])/3 - (x^3*Log[1 + (1 - I*d)*E^((2*I)*a + (2*I)*b*x)])/6
+ ((I/4)*x^2*PolyLog[2, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x))])/b - (x*PolyLog[3, -((1 - I*d)*E^((2*I)*a + (2*I
)*b*x))])/(4*b^2) - ((I/8)*PolyLog[4, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x))])/b^3

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Rubi [A]  time = 0.304345, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {6264, 2184, 2190, 2531, 6609, 2282, 6589} \[ -\frac{x \text{PolyLog}\left (3,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{i \text{PolyLog}\left (4,-(1-i d) e^{2 i a+2 i b x}\right )}{8 b^3}+\frac{i x^2 \text{PolyLog}\left (2,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)+\frac{1}{12} i b x^4 \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCoth[1 - I*d + d*Tan[a + b*x]],x]

[Out]

(I/12)*b*x^4 + (x^3*ArcCoth[1 - I*d + d*Tan[a + b*x]])/3 - (x^3*Log[1 + (1 - I*d)*E^((2*I)*a + (2*I)*b*x)])/6
+ ((I/4)*x^2*PolyLog[2, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x))])/b - (x*PolyLog[3, -((1 - I*d)*E^((2*I)*a + (2*I
)*b*x))])/(4*b^2) - ((I/8)*PolyLog[4, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x))])/b^3

Rule 6264

Int[ArcCoth[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcCoth[c + d*Tan[a + b*x]])/(f*(m + 1)), x] + Dist[(I*b)/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c + I*d +
c*E^(2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c + I*d)^2, 1]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x^2 \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx &=\frac{1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))+\frac{1}{3} (i b) \int \frac{x^3}{1+(1-i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac{1}{12} i b x^4+\frac{1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{3} (b (i+d)) \int \frac{e^{2 i a+2 i b x} x^3}{1+(1-i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac{1}{12} i b x^4+\frac{1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{1}{2} \int x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right ) \, dx\\ &=\frac{1}{12} i b x^4+\frac{1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{i x^2 \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{i \int x \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right ) \, dx}{2 b}\\ &=\frac{1}{12} i b x^4+\frac{1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{i x^2 \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{x \text{Li}_3\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}+\frac{\int \text{Li}_3\left ((-1+i d) e^{2 i a+2 i b x}\right ) \, dx}{4 b^2}\\ &=\frac{1}{12} i b x^4+\frac{1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{i x^2 \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{x \text{Li}_3\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_3(i (i+d) x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac{1}{12} i b x^4+\frac{1}{3} x^3 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{i x^2 \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{x \text{Li}_3\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{i \text{Li}_4\left (-(1-i d) e^{2 i a+2 i b x}\right )}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.187783, size = 155, normalized size = 0.91 \[ \frac{1}{3} x^3 \coth ^{-1}(d \tan (a+b x)-i d+1)-\frac{6 i b^2 x^2 \text{PolyLog}\left (2,-\frac{i e^{-2 i (a+b x)}}{d+i}\right )+6 b x \text{PolyLog}\left (3,-\frac{i e^{-2 i (a+b x)}}{d+i}\right )-3 i \text{PolyLog}\left (4,-\frac{i e^{-2 i (a+b x)}}{d+i}\right )+4 b^3 x^3 \log \left (1+\frac{i e^{-2 i (a+b x)}}{d+i}\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCoth[1 - I*d + d*Tan[a + b*x]],x]

[Out]

(x^3*ArcCoth[1 - I*d + d*Tan[a + b*x]])/3 - (4*b^3*x^3*Log[1 + I/((I + d)*E^((2*I)*(a + b*x)))] + (6*I)*b^2*x^
2*PolyLog[2, (-I)/((I + d)*E^((2*I)*(a + b*x)))] + 6*b*x*PolyLog[3, (-I)/((I + d)*E^((2*I)*(a + b*x)))] - (3*I
)*PolyLog[4, (-I)/((I + d)*E^((2*I)*(a + b*x)))])/(24*b^3)

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Maple [C]  time = 30.058, size = 2339, normalized size = 13.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccoth(1-I*d+d*tan(b*x+a)),x)

[Out]

-1/2/b^2*a^2*d/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))*x+1/2/b^2*d/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*
x*a^2-1/2/b^2*a^2*d/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))*x-1/6*x^3*ln(d)-1/12*I*x^3*Pi*csgn(I*d/(exp(
2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))*csgn(d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^2+1/12*I*b*x^4-1/8*I/b^3*d/(I+
d)*polylog(4,I*(I+d)*exp(2*I*(b*x+a)))+1/6*I/b^3*a^3/(I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)+1/3*I/b
^3/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a^3-1/4*I/b^2/(I+d)*polylog(3,I*(I+d)*exp(2*I*(b*x+a)))*x-1/2*I/b^3*a^
3/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/6*I/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x^3-1/2/b^3*a^2/(I+
d)*dilog(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/2/b^3*a^2/(I+d)*dilog(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/1
2*I*x^3*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*(b*x+a))+1))^2+1/12*I*x^3*Pi*csgn(I/(exp(
2*I*(b*x+a))+1))*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(exp(2*I*(b*x+a))+1))^2+1/8/b^3/(I+d)*polylo
g(4,I*(I+d)*exp(2*I*(b*x+a)))-1/3*x^3*ln(exp(I*(b*x+a)))+1/6*x^3*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)+1
/4*I/b*d/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x+a)))*x^2+1/12*I*x^3*Pi*csgn(I*d)*csgn(I*exp(2*I*(b*x+a))/(exp(2*
I*(b*x+a))+1))*csgn(I*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))+1/12*I*x^3*Pi*csgn(I*exp(I*(b*x+a)))^2*csgn(I*e
xp(2*I*(b*x+a)))-1/6*I*x^3*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a)))^2-1/12*I*x^3*Pi*csgn(I*d)*csgn(I
*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^2-1/12*I*x^3*Pi*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*(b*x+a))+1))*csgn(I
*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^2-1/12*I*x^3*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*exp(2*I*(b*x+a))
/(exp(2*I*(b*x+a))+1))^2+1/2*I/b^3*a^2*d/(I+d)*dilog(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+1/2*I/b^3*a^2*d/(I+d
)*dilog(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+1/2*I/b^2/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x*a^2-1/4/b^2*d/(I
+d)*polylog(3,I*(I+d)*exp(2*I*(b*x+a)))*x-1/2/b^3*a^3*d/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/2/b^3*
a^3*d/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+1/6/b^3*a^3*d/(I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))
*d+I)+1/3/b^3*d/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a^3+1/12*I*x^3*Pi*csgn(I*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(
b*x+a)))^3-1/12*I*x^3*Pi*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(exp(2*I*(b*x+a))+1))^3-1/12*I*x^3*P
i*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(exp(2*I*(b*x+a))+1))*csgn((I*exp(2*I*(b*x+a))+exp(2*I*(b*x
+a))*d+I)/(exp(2*I*(b*x+a))+1))+1/12*I*x^3*Pi*csgn(I*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))*csgn(d/(exp(2*I*
(b*x+a))+1)*exp(2*I*(b*x+a)))-1/4*I/b^3*d/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x+a)))*a^2-1/2*I/b^2*a^2/(I+d)*ln
(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))*x-1/2*I/b^3*a^3/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+1/12*I*x^3
*Pi*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I))*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(exp(2*
I*(b*x+a))+1))^2+1/12*I*x^3*Pi*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(exp(2*I*(b*x+a))+1))*csgn((I*
exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(exp(2*I*(b*x+a))+1))^2-1/2*I/b^2*a^2/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(
I+d))^(1/2))*x-1/4/b/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x+a)))*x^2+1/4/b^3/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*
x+a)))*a^2-1/6*d/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x^3+1/12*I*x^3*Pi*csgn((I*exp(2*I*(b*x+a))+exp(2*I*(b*x+
a))*d+I)/(exp(2*I*(b*x+a))+1))^2-1/12*I*x^3*Pi*csgn(d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^2+1/12*I*x^3*Pi*c
sgn(d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^3-1/12*I*x^3*Pi*csgn((I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(e
xp(2*I*(b*x+a))+1))^3+1/12*I*x^3*Pi*csgn(I*exp(2*I*(b*x+a)))^3+1/12*I*x^3*Pi*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*
(b*x+a))+1))^3-1/12*I*x^3*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I))*cs
gn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(exp(2*I*(b*x+a))+1))+1/12*I*x^3*Pi*csgn(I/(exp(2*I*(b*x+a))+1)
)*csgn(I*exp(2*I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*(b*x+a))+1))

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Maxima [B]  time = 1.2286, size = 460, normalized size = 2.71 \begin{align*} \frac{\frac{12 \,{\left ({\left (b x + a\right )}^{3} - 3 \,{\left (b x + a\right )}^{2} a + 3 \,{\left (b x + a\right )} a^{2}\right )} \operatorname{arcoth}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )}{b^{2}} - \frac{-3 i \,{\left (b x + a\right )}^{4} + 12 i \,{\left (b x + a\right )}^{3} a - 18 i \,{\left (b x + a\right )}^{2} a^{2} +{\left (8 i \,{\left (b x + a\right )}^{3} - 18 i \,{\left (b x + a\right )}^{2} a + 18 i \,{\left (b x + a\right )} a^{2}\right )} \arctan \left (-d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), d \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) +{\left (-12 i \,{\left (b x + a\right )}^{2} + 18 i \,{\left (b x + a\right )} a - 9 i \, a^{2}\right )}{\rm Li}_2\left ({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) +{\left (4 \,{\left (b x + a\right )}^{3} - 9 \,{\left (b x + a\right )}^{2} a + 9 \,{\left (b x + a\right )} a^{2}\right )} \log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} +{\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \,{\left (4 \, b x + a\right )}{\rm Li}_{3}({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \,{\rm Li}_{4}({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{2}}}{36 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(1-I*d+d*tan(b*x+a)),x, algorithm="maxima")

[Out]

1/36*(12*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arccoth(d*tan(b*x + a) - I*d + 1)/b^2 - (-3*I*(b*x
+ a)^4 + 12*I*(b*x + a)^3*a - 18*I*(b*x + a)^2*a^2 + (8*I*(b*x + a)^3 - 18*I*(b*x + a)^2*a + 18*I*(b*x + a)*a^
2)*arctan2(-d*cos(2*b*x + 2*a) + sin(2*b*x + 2*a), d*sin(2*b*x + 2*a) + cos(2*b*x + 2*a) + 1) + (-12*I*(b*x +
a)^2 + 18*I*(b*x + a)*a - 9*I*a^2)*dilog((I*d - 1)*e^(2*I*b*x + 2*I*a)) + (4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9
*(b*x + a)*a^2)*log((d^2 + 1)*cos(2*b*x + 2*a)^2 + (d^2 + 1)*sin(2*b*x + 2*a)^2 + 2*d*sin(2*b*x + 2*a) + 2*cos
(2*b*x + 2*a) + 1) + 3*(4*b*x + a)*polylog(3, (I*d - 1)*e^(2*I*b*x + 2*I*a)) + 6*I*polylog(4, (I*d - 1)*e^(2*I
*b*x + 2*I*a)))/b^2)/b

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Fricas [C]  time = 1.83746, size = 980, normalized size = 5.76 \begin{align*} \frac{i \, b^{4} x^{4} + 2 \, b^{3} x^{3} \log \left (\frac{{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) + 6 i \, b^{2} x^{2}{\rm Li}_2\left (\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, b^{2} x^{2}{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - i \, a^{4} + 2 \, a^{3} \log \left (\frac{{\left (2 \, d + 2 i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt{4 i \, d - 4}}{2 \, d + 2 i}\right ) + 2 \, a^{3} \log \left (\frac{{\left (2 \, d + 2 i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt{4 i \, d - 4}}{2 \, d + 2 i}\right ) - 12 \, b x{\rm polylog}\left (3, \frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \, b x{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 2 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 2 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 i \,{\rm polylog}\left (4, \frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 12 i \,{\rm polylog}\left (4, -\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(1-I*d+d*tan(b*x+a)),x, algorithm="fricas")

[Out]

1/12*(I*b^4*x^4 + 2*b^3*x^3*log(((d + I)*e^(2*I*b*x + 2*I*a) + I)*e^(-2*I*b*x - 2*I*a)/d) + 6*I*b^2*x^2*dilog(
1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) + 6*I*b^2*x^2*dilog(-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) - I*a^4 + 2*a^3
*log(((2*d + 2*I)*e^(I*b*x + I*a) + I*sqrt(4*I*d - 4))/(2*d + 2*I)) + 2*a^3*log(((2*d + 2*I)*e^(I*b*x + I*a) -
I*sqrt(4*I*d - 4))/(2*d + 2*I)) - 12*b*x*polylog(3, 1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) - 12*b*x*polylog(3,
-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) - 2*(b^3*x^3 + a^3)*log(1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a) + 1) - 2*(b^
3*x^3 + a^3)*log(-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a) + 1) - 12*I*polylog(4, 1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*
a)) - 12*I*polylog(4, -1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)))/b^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acoth(1-I*d+d*tan(b*x+a)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arcoth}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(1-I*d+d*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^2*arccoth(d*tan(b*x + a) - I*d + 1), x)